HDU 1003 动态规划DP Max Sum

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

 

     AC代码：

1. #include<cstdio>
2. #include<cmath>
3. #include<cstring>
4. #include<algorithm>
5. using namespace std;
6. int num[100010],sum[100010],start[100010],pos,ans,tot,t,n;     //sum[i]存储以i结尾的最大子序列，start[i]为其对应的第一个位置
7. int main()
8. {
9.     int cnt=1;
10.     scanf("%d",&t);
11.     while(t--){
12.         scanf("%d",&n);
13.         for(int i=0;i<n;i++)
14.             scanf("%d",&num[i]);
15.         for(int i=0;i<100010;i++)
16.             sum[i]=-1000000000;
17.         sum[0]=num[0];
18.         start[0]=0;pos=1;tot=1;
19.         for(int u=1;u<n;u++){
20.             if(sum[u-1]+num[u]>sum[u]){sum[u]=sum[u-1]+num[u];start[u]=start[u-1];}    //状态转移方程
21.             if(num[u]>sum[u]){sum[u]=num[u];start[u]=u;}
22.         }
23.         ans=-1000000000;
24.         for(int i=0;i<n;i++){
25.             if(sum[i]>ans){
26.                 ans=sum[i];
27.                 pos=start[i]+1;
28.                 tot=i+1;
29.             }
30.         }
31.         printf("Case %d:\n%d %d %d\n",cnt++,ans,pos,tot);
32.         if(t) printf("\n");
33. 
    
34. 
    
35.     }
36.     return 0;
37. }